24. Swap Nodes in Pairs

题目

Given a linked list, swap every two adjacent nodes and return its head.

For example,
Given 1->2->3->4, you should return the list as 2->1->4->3.

Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can be changed.

思路

本题是链表操作，每两个指针就互换一次，这里要注意一个小技巧就是，建立一个指针头，这样可以避免处理head的边界情况，这个是真的很重要啊，一开始自己就是因为犯懒，没指针头，导致各种if-else语句来对临界情况判断，超烦的。

代码

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* swapPairs(ListNode* head) {
    if(head == NULL)  
        return NULL;  
    ListNode* helper = new ListNode(0);  
    helper->next = head;  //辅助表头，可以避免处理head的边界情况
    ListNode* pre = helper;  
    ListNode* cur = head;  
    while(cur!=NULL && cur->next!=NULL)  
    {  
        ListNode* next = cur->next->next;  
        cur->next->next = cur;  
        pre->next = cur->next;  
        if(next!=NULL && next->next!=NULL)  
            cur->next = next->next;  
        else  
            cur->next = next;  
        pre = cur;  
        cur = next;  
    }  
    return helper->next;  
    }
};